On $X=CP^2\#k{(CP^2)}$ in $k$ generic points, let $h^i=\dim H^i(T^{1,0}X)$, for $i\ge 0$. First, we know $h^i=0$ for $i\ge 2$. By Riemann–Roch formula, I obtain that $h^0h^1 = 82k$. Would someone be kind to show me how to compute the general formula for $h^0$? Thank you in advance.
2 Answers
$H^0(X,T^{1,0}_X)$ is the space of holomorphic vector fields on $\mathbb{P}^2$ vanishing along your $k$ points. Since they are generic, and the holomorphic bundle $T_{\mathbb{P}^2}$ is generated by its global sections, they impose $2k$ independent conditions, so $h^0(T^{1,0}_X)$ is $82k$ for $k\leq 4$, and $0$ for $k\geq 5$.

$\begingroup$ I am not convinced the formula of $h^0=82k$, since $h^1$ is not necessary to be zero on certain X. $\endgroup$ Aug 26, 2014 at 21:22

1$\begingroup$ Last try : since $h^2=0$, $h^1$ is the number of moduli of $X$. 4 general points (or less) in $\mathbb{P}^2$ have no moduli. $\endgroup$– abxAug 27, 2014 at 15:47
On CP^2, assume the cover is given by $U_i=\{[z_0, z_1, z_2] z_i\ne 0\}$ where $i=0,1,2$, then one can compute that on $U_0$, $H^0(T^{1,0}CP^2)$ is spanned by $\frac{\partial}{\partial z_1}$, $\frac{\partial}{\partial z_2}$, $z_1\frac{\partial}{\partial z_1}$, $z_1\frac{\partial}{\partial z_2}$, $z_2\frac{\partial}{\partial z_1}$, $z_2\frac{\partial}{\partial z_2}$, $z_1z_2\frac{\partial}{\partial z_1}+z^2_2\frac{\partial}{\partial z_2}$, $z^2_1\frac{\partial}{\partial z_1}+z_1z_2\frac{\partial}{\partial z_2}$. If we blow up point at $[1,0,0]$, then only $z_1\frac{\partial}{\partial z_1}$, $z_1\frac{\partial}{\partial z_2}$, $z_2\frac{\partial}{\partial z_1}$, $z_2\frac{\partial}{\partial z_2}$, $z_1z_2\frac{\partial}{\partial z_1}+z^2_2\frac{\partial}{\partial z_2}$, $z^2_1\frac{\partial}{\partial z_1}+z_1z_2\frac{\partial}{\partial z_2}$ can be extended to $CP^2\#(CP^2)$. Moreover, if we continue blowing up at $[0,1,0]$, one can check that only $z_1\frac{\partial}{\partial z_1}$, $z_1\frac{\partial}{\partial z_2}$, $z_2\frac{\partial}{\partial z_2}$, $z_1z_2\frac{\partial}{\partial z_1}+z^2_2\frac{\partial}{\partial z_2}$ are extendable. Continue blowing up at $[0,0,1]$, then on $CP^2\#3(CP^2)$, the only left vector fields are $z_1\frac{\partial}{\partial z_1}$ and $z_2\frac{\partial}{\partial z_2}$. And do blow up one more time, e.x. at [1,1,1], then there is no vector fields extendable. This verifies that $h^0=82k$ when $k\le 4$ and $h^0=0$ when $k\ge 5$.